= Solution
Let $p$ first be an arbitrary degree-$n$ <polynomial> with leading coefficient $a_n\ne0$. Choose $R$ so large that on $|z|=R$,
$$
|p(z)-a_nz^n|<|a_n|R^n.
$$
The homotopy $a_nz^n+s[p(z)-a_nz^n]$, $0\le s\le1$, never meets zero on that <circle>. Hence the image curve $p(Re^{i\theta})$ has <winding number> $n$, the same as $a_nR^ne^{in\theta}$. The <argument principle> counts precisely $n$ zeros inside, with multiplicity, since there are no poles. Choose $R$ large enough that the same leading-term domination holds for every $|z|\ge R$; the lower-term ratios decrease with radius, so this also excludes all exterior zeros. This proves the <fundamental theorem of algebra> with its multiplicity count. A degree-zero nonzero <polynomial> has no zeros.
Now use the specified bound for the monic quartic. Compare $\Gamma$ with the closed curve
$$
\gamma_0(t)=\begin{cases}R^4e^{4\pi it},&0\le t\le1/2,\\R^4,&1/2\le t\le1.\end{cases}
$$
Its first piece travels once counterclockwise around a <circle>, and its second stays at the endpoint, so its <winding number> is one. On the first piece, $|\Gamma-\gamma_0|<R^4$. On the second, both $p(R)$ and $p(iR)$ differ from $R^4$ by less than $R^4$; every convex combination of those errors also has modulus less than $R^4$. Thus the quoted <winding number> stability result gives
$$
\boxed{n(\Gamma,0)=1.}
$$
For real coefficients define $q(y)=y^4-by^2+d$. It has no real zeros and positive leading coefficient, so $q(y)>0$ for every real $y$, including $d=q(0)>0$. Therefore
$$
p(iy)=q(y)+i(cy-ay^3)
$$
lies strictly in the right half-plane for every real $y$. Also $p(x)>0$ for $x\ge0$: there are no zeros there, and the <polynomial> is positive for sufficiently large positive $x$. These facts exclude roots on the positive real and imaginary axes.
Consider the positively oriented quarter-disk boundary: first the arc from $R$ to $iR$, then the imaginary axis to zero, then the real axis back to $R$. Under $p$, its last two pieces form a path from $p(iR)$ to $p(R)$ entirely within the right half-plane. The chord closing $\Gamma$ also lies there. Since this half-plane is convex and excludes zero, deforming one closing path into the other changes no <winding number>. Thus the full image of the quarter-disk boundary has <winding number> one. By the <argument principle> there is exactly one root inside that quarter-disk, counted with multiplicity.
Finally, all <polynomial> roots lie in $|z|<R$ under the given strict coefficient bound: for $|z|\ge R$, the sum of lower terms divided by $|z|^4$ is smaller than one. Hence
$$
\boxed{\text{exactly one zero lies in the open first quadrant}.}
$$
Its multiplicity is one. This is the <quarter-sector winding test for a real quartic>; it accounts for the axis exclusions and does not replace the sector boundary by a chord without justifying the deformation.
Back to article page
